We could take the easy way out and just say
(110 kW) x (3 hours) = 330 kilowatt hours .
But that's cheap, and hardly worth even 5 points.
If we want to talk energy, let's use the actual scientific unit of energy.
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" 110 kw " means 110,000 watts = 110,000 joules/second .
(3 hours) x (3600 sec/hour) = 10,800 seconds.
(110,000 joules/second) x (10,800 seconds) = 1.188 x 10⁹ Joules
That's
==> 1,188,000,000 joules
==> 1,188,000 kilojoules
==> 1,188 megajoules
==> 1.188 gigajoules
Atsa nawfulotta energy !
It goes back to that "110 kw appliance" that we started with.
That's no common ordinary household appliance. 110 kw is something like
147 horsepower. In order to bring 110 kw into your house, you'd need to
take 458 Amperes through the 240-volt line from the pole. Most houses
are limited to 100 or 200 Amperes, tops. And the TRANSFORMER on
the pole, that supplies the whole neighborhood, is probably a 50 kw unit.
Answer:
The angle is 
Explanation:
From the question we are told that
The mass is 
The radius is 
The speed is 
According to the law of energy conservation
The potential energy of the mass at the top is equal to the kinetic energy at the bottom i.e
=> 
Here h is the vertical distance traveled by the mass which is also mathematically represented as
So
![\theta = sin ^{-1} [ \frac{1}{2* g* r } * v^2]](https://tex.z-dn.net/?f=%5Ctheta%20%20%20%3D%20sin%20%5E%7B-1%7D%20%5B%20%5Cfrac%7B1%7D%7B2%2A%20g%2A%20r%20%7D%20%2A%20%20v%5E2%5D)
substituting values
![\theta = sin ^{-1} [ \frac{1}{2* 9.8* 1.1 } * (3.57)^2]](https://tex.z-dn.net/?f=%5Ctheta%20%20%20%3D%20sin%20%5E%7B-1%7D%20%5B%20%5Cfrac%7B1%7D%7B2%2A%209.8%2A%201.1%20%7D%20%2A%20%20%283.57%29%5E2%5D)
I think it blows vertically and horizontally cause wind can blow different directions
The difference between true north and magnetic north is called "declination."
Answer:
Angle = 0.2520 radians
Explanation:
Complete question:
Sound with frequency 1220Hz leaves a room through a doorway with a width of 1.13m . At what minimum angle relative to the centerline perpendicular to the doorway will someone outside the room hear no sound? Use 344m/s for the speed of sound in air and assume that the source and listener are both far enough from the doorway for Fraunhofer diffraction to apply. You can ignore effects of reflections.
Given Data:
Speed of sound =v= 344 m/sec ;
Width of doorway =d= 1.13m ;
Frequency of sound =f= 1220 Hz ;
Solution:
As we know that
Wvelength = w = v/f = 344/1220 = 0.281967m
Now we also know that
w = dsin(A) where A is the angle
A = arcsin(w/d) =14.44° = 14.44*(3.14/180) = 0.2520 radians
At the angle of 0.252 radians relative to the centreline perpendicular to the doorway a person outside the room will hear no sound under given conditions.
